jmc

Let the two disjoint subsets be $A$ and $B$, and let $C=S-(A+B)$. For each $i\in S$, either $i\in A$, $i\in B$, or $i\in C$. So there are $3^{10}$ ways to organize the elements of $S$ into disjoint $A$, $B$, and $C$. However, there are $2^{10}$ ways to organize the elements of $S$ such that $A=\varnothing$ and $S=B+C$, and there are $2^{10}$ ways to organize the elements of $S$ such that $B=\varnothing$ and $S=A+C$. But, the combination such that $A=B=\varnothing$ and $S=C$ is counted twice. Thus, there are $3^{10}-2\cdot 2^{10}+1$ ordered pairs of sets $(A,B)$. But since the question asks for the number of unordered sets $\{A,B\}$, $n=\frac{1}{2}(3^{10}-2\cdot 2^{10}+1)=28501\equiv \boxed{501}(\mathrm{mod}1000)$.